3 and 4 .Determinants and Matrices
easy

If $\left| {\,\begin{array}{*{20}{c}}1&k&3\\3&k&{ - 2}\\2&3&{ - 1}\end{array}\,} \right| = 0$,then the value of $ k $ is

A

$-1$

B

$0$

C

$1$

D

None of these

(IIT-1979)

Solution

(d) $\left| {\,\begin{array}{*{20}{c}}1&k&3\\3&k&{ – 2}\\2&3&{ – 1}\end{array}\,} \right|\, = \,0 \Rightarrow k = \frac{{33}}{8}$.

Standard 12
Mathematics

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